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So, What Do You Do Here?

Trivial Truth Functions

A function that is trivial for some trivial domain returns the same value as a function without the trivial variables, for every condition value of the trivial function.

$$(\textbf{A}^n,\textbf{B}^m)_\alpha=(\textbf{A}^n)_\beta\ \forall\ (\textbf{A}^n,\textbf{B}^m)_\text{CV}$$

Where $\alpha$ is the trivial function and $\textbf{B}^m$ is the trivial domain.

The equation above states that the two functions will return identical values for every condition value of $(\textbf{A}^n,\textbf{B}^m)$. The equality test can be executed by a biconditional, since it will return one when both inputs are zero or both are one, exclusively. Consequently, if every condition value of $(\textbf{A}^n,\textbf{B}^m)$ causes the biconditional to return one, then the biconditional is a tautology. As such, the equation above can be tested by conjunctive neglect of a biconditional between $\alpha$ and $\beta$.

$$\Big<\big((\textbf{A}^n,\textbf{B}^m)_\alpha,(\textbf{A}^n)_\beta\big)_\mathtt{0x9}\Big>^{(\textbf{A}^n,\textbf{B}^m)}_\cap$$

The commutativity of conjunctive neglect allows the above neglect can be broken up into two nested expressions.

$$\bigg<\Big<\big((\textbf{A}^n,\textbf{B}^m)_\alpha,(\textbf{A}^n)_\beta\big)_\mathtt{0x9}\Big>^{(\textbf{B}^m)}_\cap\bigg>^{(\textbf{A}^n)}_\cap$$

Then, expanding the inner neglect reveals,

$$\bigg<\Big(\big((\textbf{A}^n,\textbf{B}^m)_\alpha\big|_{(\textbf{B}^m)_\text{CV}=0},(\textbf{A}^n)_\beta\big)_\mathtt{0x9},\ldots,\big((\textbf{A}^n,\textbf{B}^m)_\alpha\big|_{(\textbf{B}^m)_\text{CV}=2^m-1},(\textbf{A}^n)_\beta\big)_\mathtt{0x9}\Big)_\cap\bigg>^{(\textbf{A}^n)}_\cap$$

where the $\cap$, used after the truth function, signifies repeated conjunction.

This statement will only return one if every semi-parse of $\alpha$, over $\textbf{B}^m$, is equal to $\beta$ for all conditional values of $\textbf{A}^n$.

$$(\textbf{A}^n,\textbf{B}^m)_\alpha\big|_{(\textbf{B}^m)_\text{CV}=i}=(\textbf{A}^n)_\beta\ \forall\ i\in[0,2^m-1]\land\ \forall\ (\textbf{A}^n)_\text{CV}$$

And, since every semi-parse of $\alpha$ is equal to the same function, they are all equal to each other.

$$(\textbf{A}^n,\textbf{B}^m)_\alpha\big|_{(\textbf{B}^m)_\text{CV}=i}=(\textbf{A}^n,\textbf{B}^m)_\alpha\big|_{(\textbf{B}^m)_\text{CV}=j}\ \forall\ i,j\in[0,2^m-1]\land\ \forall\ (\textbf{A}^n)_\text{CV}$$

That is, a truth function, $(\textbf{A}^n,\textbf{B}^m)_\alpha$, is trivial, with respect to some domain $(\textbf{B}^m)$, if every semi-parse of $\alpha$, over $(\textbf{B}^m)$, are identical.

$$(\textbf{A}^n,\textbf{B}^m)_\alpha=(\textbf{A}^n)_\beta\iff(\textbf{A}^n,\textbf{B}^m)_\alpha\big|_{(\textbf{B}^m)_\text{CV}=i}=(\textbf{A}^n,\textbf{B}^m)_\alpha\big|_{(\textbf{B}^m)_\text{CV}=j}\ \forall i,j\in[0,2^m-1]$$
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